Answer: Law of conservation of energy
Heat Loss by the metal = Heat Gain by the water (Assume no heat loss to the environment)
M₁C₁ΔT₁= M₂C₂ΔT₂
Where
C₁ and C₂ = the specific heat capacities of the metal and water = C₁?, 4.182J/gK
ΔT₁ and ΔT₂ = Temperature changes of the metal and the water
ΔT₁ = (373 - 300.8)K and ΔT₂= (300.8 - 296.7)K convert to S.I units
M₁ and M₂ are the masses of the metal and water = 59.047g, but
M₂ = density x volume = 1kg/m³ x 0.1m³=0.1kg
Note: Volume of water = 100ml=0.1m³, also density of water = 1kg/m³
We have,
0.059kg x C₁J/KgK (373 - 300.8)K = 0.1kg x 4182J/kgK (300.8 - 296.7)K
4.259C₁=1715J/KgK
C₁=402.5J/KgK
The specific heat of the metal is 402.5J/KgK
Answer:
V = 6.55*10^{-6} v
Explanation:
The number density can be determined by using below formula:

where,
B is uniform magnetic field 0.74
i is current 18 A
V is hall potential difference
l is thickness 150 MICRO METER
e is electron charge 1.6 *10^{-19} C
therefore V can be determined as


V = 6.55*10^{-6} v
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
Answer:
vf = v₁/3 + 2v₂/3
Explanation:
Using the law of conservation of linear momentum,
momentum before impact = momentum after impact
So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.
Mv₁ + 2Mv₂ = 3Mvf
dividing through by 3M, we have
v₁/3 + 2v₂/3 = vf
vf = v₁/3 + 2v₂/3
here the answer to your question hope it helps you
electric circuit